Graded ring generated by finitely many homogeneous elements of positive degree has Veronese subring finitely generated in degree one

Question

Let $S=\bigoplus_{k\ge 0}S_n$ be a graded ring which is generated over $S_0$ by some homogeneous elements $f_1,\dotsc, f_r$ of degrees $d_1,\dotsc, d_r\ge 1$, respectively. I want to show that there exist some integer $N>0$ such that $\bigoplus_{k\ge 0}S_{kN}$ is generated over $S_0$ by $S_N$. This result is useful in algebraic geometry because it allows one to reduce oneself to the case in which $S$ is generated by $S_1$ over $S_0$. However, I can't see how to solve this problem. It would be necessary and sufficient to find a $N$ which satisfies the following elementary condition:

Let $k$ be a positive integer, $a_1,\dotsc, a_r\ge0$ integers such that $a_1d_1+\cdots+a_rd_r=kN$. Then there exist integers $0\le b_i\le a_i$ such that $b_1d_1+\cdots+b_rd_r=N$.

Supposedly taking $N=rd_1\cdots d_r$ works, but I can't show that.

Answer 1

Let $g=\prod_{i=1}^r d_i$ and let $q_i=g/d_i$. Write $a_i=k_iq_i+r_i$ for integers $k_i$ and $r_i$ with $0\leq k_i$ and $0\leq r_i< q_i$. Then $$ kN = \sum a_id_i = \sum (k_iq_i+r_i)d_i = (\sum k_ig) + (\sum r_id_i),$$ so $\sum r_id_i$ is divisible by $g$. On the other hand, as $r_i < q_i$, we have $\sum r_id_i < \sum q_id_i = rg = N$ and therefore $\sum k_iq_id_i = \sum k_ig > (k-1)N$. So we can pick $c_1,\cdots,c_r$ with $0\leq c_i\leq k_i$ and $\sum c_iq_id_i = \sum c_ig = N - \sum r_id_i$, and choosing $b_i=c_iq_i+r_i$, we win: $\sum b_id_i = N$.